"

Putting it all together

Let’s put everything we learned above into something more concise. Closure is defined just as before. Thus:

Definition 8.4 (Closure). We call a branch of a tree closed provided there are some

1\mathcal{L}_1formulas of form

YY,

¬Y\neg Yoccurring on it. We call a tree closed if all its branches are closed.

Then, we can define what syntactic entailment means in first-order logic.

Definition 8.5 (Syntactic entailment). Let

A={X1,X2,...}A=\{X_1, X_2, …\}be any set of formulas and let

YYbe any formula of

1\mathcal{L}_1. Then, the set of premises

AAsyntactically entails

YY, or

YYis a syntactic consequence of

AA(in first-order logic), iff there is a closed tree with

¬Y\neg Yat its origin, where each occurrence of a formula on the tree except

¬Y\neg Yis derived using one of the following rules:

  1. the rules

    ¬¬\neg\neg,

    \wedge,

    ¬\neg \vee,

    ¬\neg \rightarrow;

  2. the rules

    \vee,

    ¬\neg \wedge,

    \rightarrow;

  3. the rules

    \forall,

    \exists,

    ¬\neg \forall,

    ¬\neg \exists;

  4. the premise rule Pr (using

    AA).

In such cases, we write

AYA \vdash Y, and call the closed tree a proof of

YYfrom the premise set (or simply, premises)

AA. If no occurrence of a formula on the tree was introduced using the rule Pr, we say

YYis a theorem of first-order logic.

Soundness and completeness

Similar to zeroth-order logic, we still have that relative to the first-order semantics introduced above, the tableaux system introduced is sound and complete.

Proposition 8.8 (Soundness). For any set of formulas

AAand any formula

YYof

1\mathcal{L}_1, if

AYA \vdash Y, then

AYA \models Y.

Remark 8.2. In other words, if a formula

YYis syntactically entailed by the set of formulas

AA, then we can be sure that

YYis also semantically entailed by

AA.

Proposition 8.9 (Completeness). For any set of formulas

AAand any formula

YYof

1\mathcal{L}_1, if

AYA \models Y, then

AYA \vdash Y.

Remark 8.3. In other words, if a formula

YYis semantically entailed by the set of formulas

AA, then we can be sure that

YYis also syntactically entailed by

AA.

Now if we put these two together, we get a biconditional of the form:

AYiffAYA \vdash Y \text{ iff } A \models Y

Thus, syntactic and semantic entailment go hand in hand in first-order logic. If one obtains, the other obtains too!

It is easy to see how this fact entails the following:

Proposition 8.10.

A⊭YA \not\models Yiff there is no tree with

¬Y\neg Yat its origin constructed with the rules enumerated above whose branches are all closed.

Proof. By soundness and completeness, we have that

AYA \models Yiff

AYA \vdash Y, so we also have that

A⊭YA \not\models Yiff

AYA \nvdash Y. Moreover,

AYA \nvdash Ymeans that

YYis not a syntactic consequence of

AA. The formula

YYwould be a syntactic consequence of

YYif there were a closed tree with

¬Y\neg Yat its origin with the rules enumerated above whose branches are all closed. So since this is not the case, there can be no such tree. ◻

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Learning Logic Backwards Copyright © by Peter Susanszky is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.